How do you find a unit vector a)parallel to and b)normal to the graph of #f(x)# at the indicated point. #f(x)= x^3, "at" (1,1)#?

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Answer 1 · 2017-10-03T08:41:47

Unit vector patallel to function at #(1,1)# is #3/sqrt10hati-1/sqrt10hatj# and normal vector is #1/sqrt10hati+3/sqrt10hatj#

The slope of tangent at #f(x)=x^3# at #(1,1)# is #f'(1)# and hence slope of normal is #-1/(f'(1))#

Now as #f(x)=x^3#, #f'(x)=3x^2#. Slope of tangent at #(1,1)# is #3#

and slope of normal is #-1/3#

Therefore equation of tangent is#y-1=3(x-1)# or #3x-y=2#

and unit vector parallel to #f(x)# would be #(3hati-hatj)/sqrt(1^+3^2)#

or #3/sqrt10hati-1/sqrt10hatj#

and equation of normal is #y-1=-1/3(x-1)#

i.e. #3y-3=-x+1# or #x+3y=4#

and unit normal vector would be #(hati+3hatj)/sqrt(1^+3^2)#

or #1/sqrt10hati+3/sqrt10hatj#, where #hati# and #hatj# are unit vectors along #x#-axis and #y#-axis respectively.